 So, let us begin today with the discussion of the quiz that we had last week and go through the answers quickly. So, you can check against your answer books we start with question one which was just true or false and the first statement was a conservative dynamical system given by an equation of motion of the form x dot is f of x x dot is f of x and we told it is conservative cannot have any attractors the answer is that it is true because for a conservative system this implies that del dot f is equal to 0 and that is so then we know that the volume element in phase space is preserved under the flow. So, since we have defined a conservative system as one in which the volume elements do not change under the flow a conservative then dynamical system cannot have any attractors because if you have an attractor no matter what the nature of the attractor either a point attractor like a critical point or a limit cycle a stable limit cycle or more complicated attractors the fact is that all the points in the basin of attraction of this attractor would eventually asymptotically fall into the attractor and therefore this definite reduction in phase space volume and since a conservative system the flow does not allow for any change in the phase space volume it cannot have any attractors Hamiltonian systems of course are special cases of conservative systems and they would not have any attractors either the next question asked the harmonic oscillators the only system whose time period of oscillation is independent of the amplitude of oscillation and this is false we have ourselves seen a number of oscillators isochronous oscillators whose time periods are actually independent of the amplitude of oscillation these are called isochronous oscillators because it means that the amplitude the time period of oscillation is independent of the energy of the oscillator under special circumstances this can happen even in a non-linear problem yes I stated that a basic characteristic of the linear harmonic oscillator is that it is isochronous that it does not have a time period which is dependent on the amplitude or the energy but the converse is not necessarily true that if you have isochronicity it does not necessarily imply that you have a harmonic oscillator you could have non-linear oscillators which would do the same thing here is an example of an oscillator whose time period would actually be independent of the energy of the oscillator consider for instance motion in one dimension with a Hamiltonian given by p2 over 2m plus one half m let us call it k k x2 plus a force which goes like a some constant a over 2 x2 potential of this kind now what would the force on this particle be it would be minus the derivative of the potential so the force on the particle is minus dv over dx this is equal to minus kx that's the harmonic oscillator part and then the derivative of this potential which would go like plus or x3 and what does this do as x goes to 0 what does the potential look like in this case what's the shape of the potential so let's plot v of x is equal to one half k x2 plus a over 2 x2 where k and a are positive constants and then if I plot this potential here's x versus v of x it clearly goes to infinity as x goes to 0 on both sides it also goes to infinity as x goes to plus or minus infinity and of course it's clear that in between it comes down and does this and symmetrically so on this side as well and the particle can oscillate either in this well here or in this well here quite symmetrically for the same energy could either have a center of oscillation on the left hand side or in the right hand side and if you compute the time period of oscillation in this potential it turns out that it's actually independent of where you are of what the energy is I leave that to you as an exercise to check out that this oscillator is isochronous and it's not a linear harmonic oscillator because the force is not just proportional to negative displacement but also has an extra term here when I say linear harmonic or harmonic I mean that the potential is a quadratic function of the displacement that's all that's needed anything beyond that is nonlinear and the reason it's nonlinear is because the equation of motion becomes nonlinear recall that the equation of motion here is x dot is ? h over ? p so it's p over m but p dot is – ? h over ? x that's – kx if you restricted yourself to just this but if you also included the nonlinear term you have a over x cubed so this is the reason why we call these oscillators nonlinear because the equation of motion is no longer linear it has nonlinear terms in the coordinates yeah now the reason I used I kept saying linear harmonic oscillator which you're used to calling as a simple harmonic oscillator because we also looked at the word linear there was a bit of a misnomer I should have said one dimensional harmonic oscillator because the oscillation was in one direction the x direction alone we also looked at combinations of oscillators oscillator on a plane oscillator in three dimensions and so on they're not linear harmonic oscillators they're just harmonic oscillators so I use the word linear in the other sense namely along a line motion along a line a given line should perhaps not do that should just call it a one dimensional harmonic oscillator in that case but every harmonic oscillator has a linear equation of motion linear in the coordinates and momentum and when you introduce nonlinearity in the potential you at once have these extra terms and it's not at all guaranteed that the oscillations would have a time period independent of the energy but in this particular case it turns out that they do and that's an interesting exercise to look at there are profound implications to this this model here is not pulled out of the hat it has further implications and it's a specific it's a member of a specific family of potentials a model which has many many other interesting properties so even though the potential doesn't look parabolic at all the oscillations in this are independent of the energy of the oscillator this could very well happen we saw however that if you took a potential of the form a power of x here in one dimension just a power of x no extra terms when the only oscillator which had a time period independent of the amplitude was in fact the harmonic oscillator the quadratic power here nothing else but that was restricted to the class of potentials it just had a single power of x and nothing more than that more complicated functions could do this there are in fact an infinite number of potentials which would lead to oscillations independent of the amplitude of oscillation of the energy in general okay the next question in what sense the question is is there anything which characterizes the functions and the potentials not so simple to classify not so simple it's possible to a certain extent but not so simple these models the one with the x squared and the one over x squared they form they're part of a much bigger family of potentials where this property would be satisfied not just in one dimension but in higher dimensions as well or not just with one oscillator but many oscillators on a line it's not a this is not a this is not a sufficiency condition this is not the only one that does it in the sense that it's not a it's not a unique property to this particular oscillator there are other potentials many other potentials which would do this and to some extent they can be classified we will not get into that right now the third question was straightforward it said consider a canonical transformation of an autonomous Hamiltonian system under such a transformation the form of Hamilton's equations is preserved although the functional form of the Hamiltonian in the new variables need not remain the same as the original one that in fact is what happens in a general canonical transformation so the statement is quite true the next one says every dynamical system given by an equation of the form x dot equal to f of x can be transformed into a gradient system by a suitable choice of dynamical variables so the proposition is that this could always be transformed by a suitable choice of variables to look like y dot is equal to the gradient with respect to y of phi of y no such guarantee at all because this would imply that every vector field could be writable as the gradient of some kind of scalar this is not certainly true certainly not true at all so it is not necessary that this should happen it is not true that every dynamical system can be transformed to a gradient system by suitable choice of variables change of variables we go on to the next homoclinic orbits can occur in both conservative and dissipative systems and the answer is yes indeed they can because all that a homoclinic orbit does is to start at a saddle point part of its unstable manifold goes out and comes back forms a loop eventually and the part that comes back is part of the stable manifold of the saddle point and there is no restriction on this it could happen in a dissipative system it could happen in a conservative system as well pardon me it does not matter whether the volume elements change or not because you only talk about a single trajectory here and there is no reason at all why this cannot happen independent of what else happens anywhere else linear stability analysis need not reveal the correct nature of the flow in the vicinity of a critical point that has a center manifold this is the whole point about center manifolds and it is certainly a true statement it could but on the other hand it might let you down therefore you have to go beyond linear stability analysis as we saw with specific examples once you have a center manifold the live will Arnold criterion for integrability is applicable to any even dimensional dynamical system and that is false because the live will Arnold criterion is specific to Hamiltonian systems which form a special class of dynamical systems of even dimensionality. So there are many many other systems innumerable systems which are nothing to do with Hamiltonian systems and there is no question of anything like this criterion in those cases the next statement said a bifurcation occurs at some value of a parameter in a dynamical system if the nature of the flow changes qualitatively as the parameter crosses that value and yes indeed that is the very definition that we use for elementary bifurcations the critical points corresponding to an undamped simple pendulum can only be centers of saddle points that is also true because this is a Hamiltonian system undamped simple pendulum therefore there is no attractor in the problem it is a Hamiltonian system and the only critical points you could have this simple system or centers or saddle points and we saw that you had centers and saddle points alternating corresponding to the minima and the maxima of the cosine potential consider the two dimensional dynamical system given by the following equations x square x dot is x squared minus y squared and y dot is equal to 2 x y and the proposition was the critical point at the origin is a saddle point and that is false because the saddle point a saddle point has a winding number the vector field corresponding to a saddle point has a winding number of minus 1 but this is a dipole field and the winding number here is 2 and this is topologically quite distinct from a saddle point and it is a higher order critical point because this is not even linearizable in the vicinity of the origin there are no linear terms here at all it is intrinsically nonlinear and it is happened by the coalescence of two singularities two simple critical points yes yes it is not no no it is a critical point I mean that is it so there is no question of any unfolding or anything like that this is just a critical point as it stands yes it could like a saddle note for example it could it could yes but I would not call no no I would not call it the saddle point at all because this is a the statement being made has to do with the singularity of this vector field at the origin so it is a local statement about the singularity at the origin and that is in this case undoubtedly a dipole singularity and therefore has a winding number 2 and it is not a saddle point it cannot be deformed smoothly into a saddle point because the winding number is a topological invariant and it says that no matter how you transform coordinates and shift and bend and so on you cannot change the nature of the you cannot change the winding number so a saddle point remains a saddle point on the other hand that was not true for nodes you saw for instance that something that looks like a source radial field could become a tangential field so something that looks like a node could transform into a spiral point and so on these could be done by smooth distortions but certainly you cannot take a saddle point and convert it and distort it into dipole field or anything like that so a hope bifurcation can only occur in a dissipative system and it is true because a hope bifurcation is one where a limit cycle is involved and it is a bifurcation where a stable critical point bifurcates into a stable limit cycle and an unstable critical point or if it is a subcritical bifurcation an unstable critical point bifurcates into an unstable limit cycle and a stable critical point neither case attractors are involved and therefore this sort of thing cannot happen in a Hamiltonian system or more generally in a conservative system but it can certainly and does frequently happen in dissipative systems finally if the Poisson bracket of a with b vanishes and that of b with c vanishes then the Poisson bracket of a with c necessarily vanishes that's false because all you can say from the acoby identity is if a with b is 0 and b with c is 0 this would imply when and you are asked to find out if the Poisson bracket of a with c vanishes what you can say assert is that the Poisson bracket of this is 0 by the acoby identity because the other two terms drop out and all you can say is that this quantity need be 0 at some function whose Poisson bracket with b happens to be 0 and that's about it so you can't say anything more however if you have one degree of freedom systems and their Hamiltonian systems for example a single one degree of freedom Hamiltonian system and then you know that the Hamiltonian in an autonomous case is the only functionally independent constant of the motion in the problem then anything else which you find which is also a constant of the motion would necessarily have to be a function of the Hamiltonian and then of course if a is the Hamiltonian b and c are functions of the Hamiltonian then of course all the Poisson brackets of these quantities with the Hamiltonian vanish but that's not true in general in general this is all that you can assert and we had an implication to this the implication was that in a Hamiltonian system if you set b equal to the Hamiltonian for instance then a with b equal to 0 implies that a is a constant of the motion similarly c with the Hamiltonian equal to 0 implies that c is a constant of the motion and then this statement implies that the commutator the Poisson commutator of a with c the Poisson bracket of a with c is also a constant of the motion so what it implies is that in a Hamiltonian system if you find two constants of the motion their Poisson bracket if it's not trivial is also a constant of the motion so it has a practical use in that in that instance the next question was multiple choice means you have considered a general Hamiltonian system in the first statement was the Hamiltonian is always a sum of a kinetic energy term and a potential energy term which depends only on the generalized coordinates this need not be so at all because I pointed out early on that all that a Hamiltonian system needs is that you have an even dimensional phase space with the certain structure the Poisson bracket structure canonical Poisson bracket structure and a Hamiltonian function specified to you which then determines the equations of motion of all the variables there's no restriction that the Hamiltonian should be of a form of a kinetic energy plus a potential energy that's only true for simple mechanical systems not true in general in fact even the statement I made about a Hamiltonian system namely that it should be even dimensional and it should have this canonical Poisson bracket structure could be generalized there are more general forms of writing Hamiltonian systems where you don't even need to have an even dimensionality of the space where the meaning of the Poisson bracket itself could be generalized further but that's the mathematical detail we haven't got into but certainly doesn't have to be a sum of a potential in a kinetic energy saddle node bifurcations cannot occur in this system they certainly can saddle node bifurcations we saw with an example of a potential itself that a saddle node bifurcation certainly can occur in a simple potential problem Hamiltonian problem the dynamical symmetry group of transformations need not necessarily be identical to the group of canonical transformations and that's certainly true because the dynamical symmetry group of a Hamiltonian system could be much smaller than the group of canonical transformations if you recall in n degrees of freedom the group of canonical transformations was the symplectic group 2n over the reals whereas the dynamical symmetry group would depend on whether the Hamiltonian had some special symmetries or not and most gen most Hamiltonians don't and when they do they have much smaller symmetry groups the example we took was the two dimensional harmonic oscillator which had a symmetry group which whose canonical group of canonical transformations was the symplectic group for sp4 on the reals on the other hand the symmetry group of the Hamiltonian itself was the group of rotations in four dimensions SO4 and the intersection of these two was a much smaller group which was isomorphic to SU2 so we saw that this needn't be true at all action angle variables necessarily exist for the system again no because the Hamiltonian need be integrable completely at all in fact you could have a few action variables less than n in number and that's not sufficient to integrate the system completely so the needn't exist at all in this sense remember that once an action angle pair exists then the angle variable does not appear in the Hamiltonian it becomes a cyclic coordinate and this is not always possible in general if it's fully integrable then the statement is under suitable conditions you have an action angle transformation which will then lead you to a set of variables in which all the angle variables disappear from the Hamiltonian and the Hamiltonian is a function of the action variables alone which are then constants of the motion but that needn't be true in general the next statement pertain to a general autonomous dynamical system described by a set of n coupled nonlinear first order ordinary differential equations like x dot equal to f of x the phase space can be either even dimensional or odd dimensional and that's certainly true automatically there's always at least one attractor in the system not necessary for instance if it's a conservative system then in there will need be any attractors at all the dynamics is necessarily measure preserving not at all it could be a dissipative system so it could very well have a measure which shrinks they must exist at least n functionally independent constants of the motion that do not have any time explicit time dependence no because if that happens you can't have any motion at all in an n dimensional space so if you have n constants of the motion which are independent and don't depend on time at all then the system can't even be integrated there's no motion there's absolutely nothing to do once you've specified the initial conditions the system just remains there you can't move out after that so that's not valid either we go on to question 2 it was fairly straight forward and the system was specified in polar coordinates by sin pi over r and theta dot equal to r in plane polar coordinates we were asked to find in this case the limit cycles of the system if any as well as the stability and it's immediately clear from here that limit cycles at r equal to 1 over n where n equal to 1 2 3 etc exist and then this becomes sin pi and it vanishes so r doesn't change on those points and you get circular limit cycles so you get a family of concentric limit cycles of radius one a half a third and so on and so forth now if I took the system near one of these limit cycles in the vicinity of these one of these limit cycles then r dot is approximately equal to sin pi n which is 0 plus r minus 1 over n the derivative of this at r equal to 1 over n the derivative of this is of course pi times cos pi over r but r is 1 over n so this becomes cos pi n times the derivative of this which is minus 1 over r squared which becomes minus n squared plus higher orders so this becomes equal to minus pi n squared minus 1 to the power n r minus 1 over n plus higher order terms and if n is equal to 1 this gives you a minus sign along with this and then it says r dot is proportional to r minus 1 with a positive coefficient here therefore for sufficiently large values of r this flows out because r dot is positive it goes away and if r is less than 1 it flows away inwards so it is clear immediately that at r equal to 1 this particular limit cycle this limit cycle here whatever is inside here flows out whatever starts here flows out because theta dot is r the bar is greater than 1 theta dot is a positive number and it increases and therefore you expect it flows off in this fashion and since theta dot is positive the flow is in the counterclockwise direction on this limit cycle and anything which starts in is going to flow away from it towards the next limit cycle which is the limit cycle at r is equal to a half so things flow in towards this and similarly between one third and half things flow out towards the half so you have an infinite number of limit cycles nested within each other the outermost one r equal to 1 is unstable the next one is stable the one inside is unstable and so on all the way so that is a full phase portrait what happens at r equal to 0 in this case what can you say about r equal to 0 well this function does not have a limit as r goes to 0 here that is quite clear and what you have is an accumulation of limit cycles of alternating stability so it is a crazy singular point but it is just an accumulation point for limit cycles that is about all you can say is r goes to 0 so the flow gets more and more and more intricate as you get inwards towards this it is not a simple critical point by any means it is like asking as n tends to infinity is it even or odd right so it is the same problem as before the limit does not exist the limit does not exist so the whole point is that the limit as r goes to 0 of sin pi over r does not exist there is no definite limit in what sense well it is clear that everything we write down here mathematically is modeling some physical system to some degree of accuracy so the question of you know whether it actually is described does it actually describe a physical system write down to r equal to 0 is a more point in that sense very unlikely to happen let us go on to the next topic and I would like to introduce to you the idea of Lyapunov functions Lyapunov's direct method which I briefly mentioned a little earlier and what I intend to do is to work out a little bit of this in terms of examples and maybe give some problems so that you could work out things and see how this method works now this method is useful it is a method for analyzing the stability of a critical point and it differs from the linearization method which we learnt about so far where if you recall we took a particular critical point we linearize the system about this critical point and then if the eigenvalues of the acrobin matrix at this point had no zero real part and the point was hyperbolic then we identified the stability or otherwise of this critical point based on what the real parts of these eigenvalues did if all the real parts with the real if at least one real part is positive you ended up with something that was unstable some direction which things would flow away but if all of them are negative things things flowed in asymptotically and you had asymptotic stability we also saw that if you have a center manifold if there are eigenvalues whose real parts are zero then either you have a center or you have more complicated behavior but the stability is not uniquely decided by linearization about that point in those cases Lyapunov's direct method helps you to do this and the method works as follows suppose for this dynamical system x dot equal to f of x suppose you have a critical point a critical point at x equal to zero so let's consider the origin to be a critical point without loss of generality and see what happens in the neighborhood of the origin so here's the origin and there's some neighborhood of the origin in which if I can find a function and let's call it v of x it's called a Lyapunov function with the following properties v of x is such that v of zero is equal to zero so it vanishes at the origin at the critical point and v of x is greater than zero at all other points in this neighborhood I then say v of x is a positive definite function so if this is true I say v is positive definite on the other hand if it's also possibly equal to zero at one or more points in the neighborhood other than the origin then I say it's positive semi definite and the same similar statement is true for negative definite and negative semi definite if it's less than zero everywhere then it's negative definite less than or equal to zero it's negative semi definite now v has nothing to do with the dynamical system it's just a Lyapunov function an auxiliary function which I'm going to try to find then the statements are as follows Lyapunov stability theorems and there are many of them but in the simplest form the stability statements are as follows statements are in brief one if there exists a positive definite if v dot is less than zero in the neighborhood in which v is positive definite by dot I mean the time derivative of this v then the critical point is asymptotically stable to a little less rigorous a little less stable. If there exists a positive definite v and if v dot is less than or equal to zero could vanish at some points in the neighborhood then the critical point is stable recall again that a center was stable but not asymptotically stable a spiral point an asymptotically stable spiral point definitely things fell into this spiral point but need not be stable. So these two don't exclude each other in some sense they are independent concepts and finally yes not necessarily we will see we will see examples so things could be stable and asymptotically stable but they could be asymptotically stable without being stable so the statement is got to do with finding specific Lyapunov functions because as you can see if I can find this if I can find this then definitely I assert that the critical point is asymptotically stable but if I can only prove this and there are points where it vanishes and I cannot prove that it's actually nonzero at every point that v dot is less than or at some points equal to zero then all you can say is that the critical point is stable we are going to see we are going to see this with examples right away what happens and three you also have a statement of instability if there exists once again positive definite v and v dot is greater than zero then the cp is unstable many refinements of these statements are possible but I am just giving the simplest version here and we are going to look at examples of course you could turn this around and instead of v you consider minus v then of course if v is positive definite and v dot is greater than zero it would translate into saying if v is negative definite and v dot is less than zero so instead of v you could always choose minus v as your Lyapunov function and then whatever you say about positive definiteness becomes a statement about negative definiteness now let's look at examples and see how to apply this and I am going to go through a series of examples but you will see how powerful this theorem is but we need to know what v dot has to do with things and the reason is the explanation is very simple if I consider v as a function of x and I consider dv over dt which is v dot what's this equal to well it's a function of x and therefore this is equal to ? v over ? x i x i dot but from the flow equations on the solution trajectories this is also equal to ? v over ? x i fi of x the ith component of the vector field f and this is summed over i the summation over repeated indices is implied I haven't written it down but this is i equal to 1 to n and this is of course equal to ? v dot f now what's the direction of gradient of v v is a function in what direction is gradient of v it's normal to the level surfaces of v and the flow specifies the direction in which the trajectory moves in phase space so you can see that this quantity here is telling you something about the relative directions of the gradient of v and the direction of the flow so it's like having a level surface and finding out of the flow is going inwards into the surface at all times in which case it has to go and hit some point or is it flowing out so in very heuristic terms that's the way in which the stability theorems emerge from a consideration of dv over dt but let's look at an example right away so fix ideas in our mind let's look at this simple two dimensional system oh and it's not restricted to two dimensions at all so this is true for an n dimensional system and that's what makes it interesting but let's look at simple two dimensional examples for the first example I'm going to look at is x dot equal to y y dot equal to minus x which is of course the harmonic oscillator all over again in suitable units and now what can you tell me about this what should I choose to be the Lyapunov function here is where there are no simple guidelines available what would one choose as the Lyapunov function in this instance well a good choice would be we know already that this is going to be a center at the origin and we know that it's stable but not asymptotically stable what would you choose as a Lyapunov function I'd like to choose something that's positive definite something which vanishes at the origin and in a neighborhood of the origin doesn't go to zero well I'd like to choose something that's got a definite sign so it must be at least quadratic pardon me I could choose x squared plus y squared so I certainly choose that if you choose v of x comma y equal to one half x squared plus y squared which you recognize is the energy of this oscillator and suitable units have set the mass and the frequency equal to one then this function is positive definite in a neighborhood of the origin in the sense we've defined it to be it's vanishes at the origin and it's not zero anywhere else in a neighborhood of the origin so what's the gradient of it's components are just x and y as it stands what do you get if you combine it with this so what do you get from gradient of v dot f this is the vector field f so it says take this component multiplied with this take this component multiplied with this and add the two and you get zero therefore we go back here and ask which of these applies we have a positive definite v and we have v dot not less than zero strictly but equal to zero so certainly this criterion applies and we're guaranteed that this critical point is stable now of course if you could find some other we some other function altogether where this was valid then the critical point would also be have been proved to be asymptotically stable and we know that's not the case in the harmonic oscillator so it implies that you cannot find the positive definite function v such that v dot is zero because if you did then it would contradict what we already know about the harmonic oscillator but no doesn't say that at all why should it say that if all you can prove is that v dot is less than equal to zero then all you established is that the critical point is stable okay now it's quite yes no why do you say that why do you say that why do you say that okay now I see the confusion no if you can show that v dot is strictly less than zero not zero then you've shown that it's asymptotically stable and in such cases in such cases yes indeed you proved stability but you proved a much stronger statement as well but you don't go the other way as in this example all I've succeeded in showing with this Lyapunov function is that v dot which is grad v dot f is zero so it hits this case here I'm unable to show that it's less than zero so the statement I'm making is that had I been able to find a Lyapunov function which was positive definite and for which v dot was strictly negative then I would have been led to the conclusion that this critical point is in fact asymptotically stable in addition to being stable no it needn't be but the Lyapunov criterion is telling you see it's a question of the choice of the Lyapunov function the same function doesn't satisfy this as well as that that's quite clear so let's go over this again I guess and pull out of the top of my hat a certain Lyapunov function and for this Lyapunov function I show that it's positive definite that's trivial and I see that v dot is strictly zero and then I look down here and ask which of these applies and it's this case that applies I conclude that this critical point is stable now if I didn't know anything more about the system this is all I could conclude about it and then I might wonder perhaps this critical point is not only stable but also asymptotically stable I'd like to examine if that is so if so I need to find another Lyapunov function some other cleverer choice of function where I could actually establish this and I'm unable to do so so the point about Lyapunov stability has to do with the clever choice of a Lyapunov function and the remarkable statement is if you can find even one Lyapunov function which satisfies the conditions of this theorem then you can conclude whatever the theorem states but that may not be the best this is certainly true but this is a statement about stability here so we have seen stability for this particular problem the simple harmonic problem the oscillator problem but we also know this problem can be explicitly solved and it's not asymptotically stable it's a Hamiltonian system here so the conclusion would be that you can't find such a V no matter how hard you try but you have to understand that this is not a it's not that you can do this in all cases you're trying to find out if in the absence of any information some statement can be made about stability and that's all the Lyapunov function does okay so let's see a few more examples then we come back and answer some of these questions yes you won't be able to find a successful Lyapunov function which would then satisfy this criterion as well yes yes indeed just as in this case we are not able to find a Lyapunov function I'm just asserting this so that this is so that you cannot find a Lyapunov function for which this is true in this instance no it's not I mean there are many many extensions to this but it's a starting point so let's look at the next instance the next instance we added something to the oscillator and we did the following we put minus x times if you recall x squared plus y squared but let's be general and let's put some phi of x, y and minus y phi of x, y where phi is a continuous function it has continuous first partial derivatives and so on what can one say now I still choose this V and then what do I get I get gradient of V equal to x, y and therefore V dot is the gradient of V dotted with this and what does that give you so these two terms cancel but I get a minus x squared phi and then I out here I get a minus y squared times phi so this becomes minus r squared times phi of x, y now the Lyapunov function I've chosen is certainly it's the it's just the energy of the unperturbed simple harmonic oscillator it's a positive definite function vanishes at the origin nonzero everywhere outside the origin and I happen to choose this function and decide and see that V dot in this case is minus r squared phi now what can one say suppose phi is positive definite then what would you say if this is greater than 0 phi greater than 0 in a neighborhood of the origin it's positive definite then V dot is negative definite and therefore I would say the critical point is asymptotically stable on the other hand if I know that phi is less than 0 if that's the way this function is then V dot becomes positive definite V is positive definite and I can certainly assert that the critical point is unstable and this would apply so without further analysis depending on what this function does and the fact that I know such a Lyapunov function exists I am able to make this statement about the stability of the critical point so this is the power of Lyapunov's method direct method it's called Lyapunov second method or direct method because it depends on your clever choice of these functions here let's yeah yes the question is can I choose let's do the next example and see if you can choose x squared plus y squared as a universal Lyapunov function see what happens so the other problem I don't want to erase this the other problem we looked at was the simple pendulum which two had linear harmonic harmonic oscillations for sufficiently small amplitudes so let's look at that example and see what happens there we had x dot equal to y and y dot was equal to on the right hand side minus sin x for the undamped oscillator and then if you put in damping you also had a minus y in this fashion in suitable units this was the damped simple harmonic oscillator with some special choice of units for the frequency and the damping coefficient what happens now and I am going to choose V as he suggested I choose V equal to one half x squared plus y squared so this would imply that V dot is equal to x multiplied by this so this is x y plus y multiplied by this so this is minus y squared minus y sin x I am interested in seeing what happens near the origin near the critical point at the origin where I know that they are going to be small oscillations if you didn't have damping I have included damping here so what happens now this becomes equal to y times x minus sin x minus y squared and even if I say the neighborhood that I am interested in is restricted to a small neighborhood of the origin and y is approximately 0 so I neglect the quadratic term as you can see immediately that x minus sin x could have either sin therefore you finished you cannot make a statement about whether it's positive definite or negative definite and so on not able to do that this case so this is not a very good choice what would you suggest then this is not the energy of the simple pendulum at all the undamped simple pendulum so the next choice would be to say shall I choose the energy of the simple pendulum itself undamped simple pendulum after all that too is a positive definite function let's see if you can choose that as the Lyapunov function so instead of this I replace it with half y squared plus 1 minus cos x which was the potential energy and that's a positive definite function also because the least value it has is 0 at the origin and then there's a neighborhood of the origin in which it's got only positive values so what happens now the gradient of V is equal to differentiate this respect to x and what do I get here it's a vector with the following components the derivative with respect to x this gives you sin x and y so this implies immediately that V dot is y sin x minus y sin x minus y square and this cancels out and you get V dot is minus y square what can you conclude now this is the damped simple pendulum remember it's a damped simple pendulum and we're looking at what happens in the neighborhood of the origin but we already know what the origin is what kind of critical point is the origin in the damped simple pendulum it's definitely asymptotically stable if it's an under damped pendulum it's a spiral point which falls in asymptotically stable spiral point but what have we achieved here we've got a Lyapunov function which is positive definite we've got a V dot which is minus y square and what can you say about that can I apply the first one is V dot less than 0 or can I only apply the second one this function if it's negative definite then certainly I can assert it's asymptotically stable but unfortunately this vanishes not only at the origin but all along the x axis so in a neighborhood of the origin it vanishes everywhere here and therefore it's negative semi definite not negative definite this is all one can assert and therefore the conclusion is this critical point is stable but we know it's stable and it's also asymptotically stable we've proved its stability now because the existence of this Lyapunov function guarantees that this critical point is at the very least stable but it's also unstable and that takes much harder work to do it's not enough to do this you need a much better Lyapunov function than this and there exists one where you can actually show that this as a critical point is also asymptotically stable. So I hope this goes a little way in answering some of the questions that were raised namely does this imply that or does this imply this and so on as you can see if you choose a bad Lyapunov function you can make no conclusions at all you choose a reasonable Lyapunov function you get some conclusions but it could be even stronger a more stronger result could exist but we've not been able to find it because we don't have a suitable choice of Lyapunov function. So the whole thing rests with finding a suitable Lyapunov function trying to see the best possible one is found in any given circumstance or not what would happen here if I took an undamped oscillator but I wrote an equation of motion which was our generalized oscillator I stop with that example today. So if I had x dot plus g of x equal to 0 is no damping in this problem and now I give you the following properties this is 0 g of x is less than 0 for minus a less than x less than 0 greater than 0 0 less than x less than plus a so it's an undamped oscillator with some possibly non-linear function of this kind what can one say about this what would you expect happens at the origin what kind of critical point you have at the origin it's a center you're completely right it's a center there's no damping I expect some stable oscillations about the center therefore I expect to be able to find a Lyapunov function where I have this property here what would you say is a suitable choice of Lyapunov function here yet the energy of the oscillator would be a good the energy of this oscillator would be a good Lyapunov function now it's a Hamiltonian system in this case what's the energy of this oscillator this is the force here this is like the momentum wise like the momentum so what would the energy be I choose a Lyapunov function v of x, y to be equal to one half y squared that's the kinetic energy plus the potential energy which is the integral of the force with the change of sign so this thing here is equal to 0 to x that's a positive definite function as you can easily check with these properties that's a positive definite function and now you can write down what V dot does and find out whether it's stable or asymptotically stable and a simple exercise shows you that the second of the criteria would apply and this would indeed be a stable critical point at the origin let's try just one more example if I don't finish this we look at it next time x dot is equal to y and y dot equal to let's look at the non-linear oscillator the cubic oscillator that we looked at earlier the duffing oscillator or a variant of it minus x minus x cube minus gamma y so this is the friction term with the positive coefficient gamma this corresponds to motion in a potential which is got an x squared part and an x4 part and this is just the first statement that the velocity is x dot is the momentum in this side what would be a good Lyapunov function in this case the energy once again so let's try that so let's put V of x, y equal to put a question mark to see if that's the best possible choice so one half y squared plus one half x squared plus one quarter x to the power 4 that corresponds to integrating this the force here to give you the potential this potential is just a quartic potential it's not the double well potential of the duffing oscillator that would happen if I put a plus sign here in which case you'd get an inverted parabola near the origin and then potential that goes like x4 up there but it doesn't matter you can give conclusions for both these potentials independent of what the sign you choose here is so I leave you to figure out what happens in this case what can you say about this potential what can you say about this critical point if you put it in it's not very hard to see all you have to do is to take the gradient of this V compute it and multiply this and see what happens so this implies that the gradient of V is equal to x plus x cubed you can see the first component and the other component is just y so I have an x y so this says V dot is equal to x y plus x cubed y that's this times this and then minus x y minus x cubed y minus gamma y squared and of course everything cancels out and what can we say about V dot now is it negative definite or is it negative semi definite negative semi definite so once again we see that this is not good enough all it says is that this critical point is stable doesn't yet establish if it's uns if it's asymptotically stable or not it's possible to find and I'll give you next time a better Lyapunov function where you can actually establish that it's also asymptotically stable not surprisingly it will involve gamma you need to involve the constant gamma and then you can show that you have a better Lyapunov exponent which will do the trick similarly for the simple pendulum problem choose the following Lyapunov exponent and show that the system is actually also asymptotically stable for the damn simple pendulum choose the Lyapunov exponent choose this Lyapunov exponent function and we can see that it's a positive definite function vanishes only in the at the origin in its neighborhood and then show that you indeed get V dot which is negative semi definite negative definite and therefore the critical point is indeed an asymptotically stable critical point so let me stop here today.